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Loser66

  • one year ago

Let A, B \(\in M(n, \mathbb C)\)and assume B is invertible. Show that 1) \((B^{-1}AB)^k= B^{-1}A^kB\) 2) Use 1 to show \(e^{tB^{-1}AB}=B^{-1}e^{tA}B\) Please help

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  1. Loser66
    • one year ago
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    1) is easy. But I don't know how to do 2 @ganeshie8 @dan815

  2. ganeshie8
    • one year ago
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    is it given that A is diagonal matrix ?

  3. Loser66
    • one year ago
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    Nope, that is all.

  4. UsukiDoll
    • one year ago
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    invertible meaning non-singular matrix.. there's an inverse.

  5. Loser66
    • one year ago
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    nvm, I got it. :) Thanks for being here.

  6. Loser66
    • one year ago
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    for 1) \((B^{-1}AB)^k = (B^{-1}AB)(B^{-1}AB)........(B^{-1}AB)\) open parenthesis and \(BB^{-1}=I\) gives us the answer

  7. Loser66
    • one year ago
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    for 2) by definition \(e^{tA}=\sum_{k=0}^{\infty} \dfrac{t^k}{k!}A^k\)

  8. Loser66
    • one year ago
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    we replace A by \(B^{-1}AB\) then just simplify it to get the answer.

  9. ganeshie8
    • one year ago
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    \(e^{tB^{-1}AB}=\sum_{k=0}^{\infty} \dfrac{t^k}{k!}(B^{-1}AB)^k =\sum_{k=0}^{\infty} \dfrac{t^k}{k!}B^{-1}A^kB \) ?

  10. UsukiDoll
    • one year ago
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    yeah.... this proof looks familiar. Miracrown helped me with one of them.

  11. ganeshie8
    • one year ago
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    how do you conclude

  12. Loser66
    • one year ago
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    @ganeshie8 sum and t, k are constants, right? hence, we can distribute inside, between \(B^{-1}\) and A

  13. ganeshie8
    • one year ago
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    Are you saying we can take \(B^{-1}\) out ?

  14. Loser66
    • one year ago
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    Yes

  15. ganeshie8
    • one year ago
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    \(\sum_{k=0}^{\infty} \dfrac{t^k}{k!}B^{-1}A^kB =B^{-1}\left(\sum_{k=0}^{\infty} \dfrac{t^k}{k!}A^k\right)B \) ? this looks ok intuitively but idk what you're gonna use as justification

  16. Loser66
    • one year ago
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    Look at the definition of \(e^{tA}\) above

  17. ganeshie8
    • one year ago
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    Yes looked.

  18. Loser66
    • one year ago
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    And we done, right?

  19. ganeshie8
    • one year ago
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    I am asking the justification for that distribution

  20. ganeshie8
    • one year ago
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    I think its not so obvious, at least for me it is not.

  21. Loser66
    • one year ago
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    https://en.wikipedia.org/wiki/Matrix_multiplication

  22. dan815
    • one year ago
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    |dw:1437277499951:dw|

  23. dan815
    • one year ago
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    |dw:1437277542960:dw|

  24. dan815
    • one year ago
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    |dw:1437277640041:dw|

  25. dan815
    • one year ago
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    therefore true

  26. dan815
    • one year ago
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    if u expand that summation u can show these 2 summations are equation therefore

  27. dan815
    • one year ago
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    |dw:1437277900211:dw|

  28. Loser66
    • one year ago
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    hey, barbecue, @ganeshie8 asked whether we can put the sum between B^-1 and A or not. You proved \(A(B_1+B_2) = AB_1 +AB_2\) which is irrelevant to the question. ha!!

  29. dan815
    • one year ago
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    this is what we need

  30. dan815
    • one year ago
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    \[\sum_{k=0}^{\infty} \dfrac{t^k}{k!}B^{-1}A^kB =B^{-1}\left(\sum_{k=0}^{\infty} \dfrac{t^k}{k!}A^k\right)B\]

  31. dan815
    • one year ago
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    |dw:1437278072521:dw|

  32. dan815
    • one year ago
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    you see?

  33. Loser66
    • one year ago
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    Oh, yeah, barbecue!! Thanks for making it clear.

  34. dan815
    • one year ago
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    sure thing

  35. anonymous
    • one year ago
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    consider \(C=B^{-1}AB\). now suppose \(C^{n-1}=B^{-1}A^{n-1}B\). it follows that $$C^n=C\cdot C^{n-1}=B^{-1}ABB^{-1}A^{n-1}B=B^{-1}AA^{n-1}B=B^{-1}A^nB$$. so by induction it follows \(C^n=B^{-1}A^n B\) define \(e^X=\sum_{n=0}^\infty \frac1{n!}X^n\) in analogy with the real exponential. $$\begin{align*}e^{B^{-1}AB}=\sum_{n=0}^\infty\frac1{n!}(B^{-1}AB)&=\sum_{n=0}^\infty\frac1{n!}B^{-1}A^nB\\&=B^{-1}\sum_{n=0}^\infty\frac1{n!}A^n B\\&=B^{-1}\left(\sum_{n=0}^\infty\frac1{n!}A^n\right)B\\&=B^{-1}e^AB\end{align*}$$

  36. anonymous
    • one year ago
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    the above follows because matrix multiplication distributes over addition: $$B^{-1}(A_1+A_2)B=B^{-1}(A_1B+A_2B)=B^{-1}A_1B+B^{-1}A_2B$$

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